Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-t^2 + 10t}{t^3 - 17t^2 + 70t} \times \dfrac{t^2 - 3t - 70}{-5t - 35} $
Solution: First factor out any common factors. $r = \dfrac{-t(t - 10)}{t(t^2 - 17t + 70)} \times \dfrac{t^2 - 3t - 70}{-5(t + 7)} $ Then factor the quadratic expressions. $r = \dfrac {-t(t - 10)} {t(t - 10)(t - 7)} \times \dfrac {(t - 10)(t + 7)} {-5(t + 7)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {-t(t - 10) \times (t - 10)(t + 7) } { t(t - 10)(t - 7) \times -5(t + 7)} $ $r = \dfrac {-t(t - 10)(t + 7)(t - 10)} {-5t(t - 10)(t - 7)(t + 7)} $ Notice that $(t - 10)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-t\cancel{(t - 10)}(t + 7)(t - 10)} {-5t\cancel{(t - 10)}(t - 7)(t + 7)} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $r = \dfrac {-t\cancel{(t - 10)}\cancel{(t + 7)}(t - 10)} {-5t\cancel{(t - 10)}(t - 7)\cancel{(t + 7)}} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $r = \dfrac {-t(t - 10)} {-5t(t - 7)} $ $ r = \dfrac{t - 10}{5(t - 7)}; t \neq 10; t \neq -7 $